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Re: M31-22 [#permalink]
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HunterJ wrote:
"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."



Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?


1. Any integer in even power is a perfect square.
2. \(z^4\) is a perfect square. A perfect square has odd number of factors. Since \(z^4\) equals to \(x^3*y^5\), then \(x^3*y^5\) (the product of x^3 and y^5, not the multiples separately) must also have odd number of factors.

Finding the Number of Factors of an Integer:

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

BACK TO YOUR QUESTION:
In case of a perfect square, say for \(6^2 = 2^2*3^2\), the powers of primes will be always even, so when counting the number of factors we'd have \((even + 1)(even + 1) = (2 + 1)(2 + 1) = 3*3 = odd*odd = odd\).

Tips about the perfect square:
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

Hope it helps.
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Re: M31-22 [#permalink]
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z cannot be a prefect square unless x=y. x and y are prime numbers, we cant get a prefect square with unequal powers of prime numbers. hence, x=y.
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Re: M31-22 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M31-22 [#permalink]
Bunuel wrote:
Official Solution:

For prime numbers \(x\) and \(y\), \(x^3*y^5=z^4\). How many positive factors does positive integer \(z\) have?

A. One
B. Two
C. Three
D. Four
E. Five


Notice that \(z^4\) is a perfect square, thus it must have odd number of factors. Since \(z^4\) equals to \(x^3*y^5\), then \(x^3*y^5\) must also have odd number of factors.

Now, if \(x\) and \(y\) are different primes, then the number of factors of \(x^3*y^5\) will be \((3 + 1)(5 + 1) = 24\), which is not odd. Therefore \(x\) and \(y\) MUST be the same prime.

In this case \(x^3*y^5 = x^3*x^5 = x^8 = z^4\).

\(x^8 = z^4\);

\(x^2 = z\).

Since \(x\) is prime, then the number of factors of \(z\) is \((2 + 1) = 3\).


Answer: C



SO Bunuel just to reiterate last line. Does that mean the factors are 1,x,z ?
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Re: M31-22 [#permalink]
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mbsingh wrote:
Bunuel wrote:
Official Solution:

For prime numbers \(x\) and \(y\), \(x^3*y^5=z^4\). How many positive factors does positive integer \(z\) have?

A. One
B. Two
C. Three
D. Four
E. Five


Notice that \(z^4\) is a perfect square, thus it must have odd number of factors. Since \(z^4\) equals to \(x^3*y^5\), then \(x^3*y^5\) must also have odd number of factors.

Now, if \(x\) and \(y\) are different primes, then the number of factors of \(x^3*y^5\) will be \((3 + 1)(5 + 1) = 24\), which is not odd. Therefore \(x\) and \(y\) MUST be the same prime.

In this case \(x^3*y^5 = x^3*x^5 = x^8 = z^4\).

\(x^8 = z^4\);

\(x^2 = z\).

Since \(x\) is prime, then the number of factors of \(z\) is \((2 + 1) = 3\).


Answer: C



SO Bunuel just to reiterate last line. Does that mean the factors are 1,x,z ?


Absolutely. Say if x = 2, then y = 4 and the factors of 4 are 1, 2, and 4.
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Re: M31-22 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M31-22 [#permalink]
This is a high quality question.
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Re: M31-22 [#permalink]
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Another approach would be to pick numbers, which worked for me...

\(x=3\) and \(y=3\), the question does not state that the prime number have to be distinct... this makes it easier to work with the same primes.

\(3^3*3^5=3^8\) so we know that \(3^8=z^4\), therefore \(9^4=z^4\) --> \(z=9\)

Factors of \(z=9\) --> 1,3,9
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Re: M31-22 [#permalink]
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I solved this Q like this:

\(x^3 * y^5 = (x*y)^3*y^2\)

Now y^2 has to be equal to x*y to make it a power of 4.

\(y^2= x*y\) leads to x=y
Therefore same prime numbers.

So \(z= (prime^2)^2\)

Therefore number of factors = 2+1.
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Re: M31-22 [#permalink]
Bunuel : IF z is a positive integer, does 'z^4' not have positive factors 1, z^2, z4, x and y?
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Re: M31-22 [#permalink]
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SHUBHAM GAUTAM wrote:
Bunuel : IF z is a positive integer, does 'z^4' not have positive factors 1, z^2, z4, x and y?


x and y must be equal (check the solution above). So, the factors of z^4 are 1, x, x^2 = z, x^3, x^4 = z^2, x^5, x^6 = z^4, x^7, and x^8 = z^4.
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Re: M31-22 [#permalink]
I think this is a high-quality question.

sveniga4 wrote:
z cannot be a prefect square unless x=y. x and y are prime numbers, we cant get a prefect square with unequal powers of prime numbers. hence, x=y.

This was my approach as well.
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Re: M31-22 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M31-22 [#permalink]
Bunuel wrote:
For prime numbers \(x\) and \(y\), \(x^3*y^5=z^4\). How many positive factors does positive integer \(z\) have?

A. One
B. Two
C. Three
D. Four
E. Five



I have a very easy and helpful approach in case if one stuck in such questions.

x and y are prime and it's nowhere given that they are distinct. We can assume both as 2.
z^4 = 2^2*2^5 = 2^8 = 4^4
z = 4
No. of factors of z = 3 (1,2 and 4)


Option C
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Re: M31-22 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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